3.2. Proofs
3.2
Proofs
Investigate!
Decide which of the following are valid proofs of the following
statement:
If ab is an even number, then a or b is even.
1. Suppose a and b are odd. That is, a 2k + 1 and b 2m + 1
for some integers k and m. Then
ab (2k + 1)(2m + 1)
4km + 2k + 2m + 1
2(2km + k + m) + 1.
Therefore ab is odd.
2. Assume that a or b is even - say it is a (the case where b is
even is identical). That is, a 2k for some integer k. Then
ab (2k)b
2(kb).
Thus ab is even.
3. Suppose that ab is even but a and b are both odd. Namely,
ab 2n, a 2k + 1 and b 2j + 1 for some integers n, k,
and j. Then
2n (2k + 1)(2j + 1)
2n 4k j + 2k + 2 j + 1
n 2k j + k + j + 0.5.
But since 2k j + k + j is an integer, this says that the integer
n is equal to a non-integer, which is impossible.
4. Let ab be an even number, say ab 2n, and a be an odd
number, say a 2k + 1.
ab (2k + 1)b
2n 2kb + b
2n − 2kb b
2(n − kb) b.
!
Therefore b must be even.
Attempt the above activity before proceeding
!
213
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3. Symbolic Logic and Proofs
Anyone who doesn’t believe there is creativity in mathematics clearly
has not tried to write proofs. Finding a way to convince the world that
a particular statement is necessarily true is a mighty undertaking and
can often be quite challenging. There is not a guaranteed path to success
in the search for proofs. For example, in the summer of 1742, a German
mathematician by the name of Christian Goldbach wondered whether every even integer greater than 2 could be written as the sum of two primes.
Centuries later, we still don’t have a proof of this apparent fact (computers have checked that “Goldbach’s Conjecture” holds for all numbers less
than 4 × 1018 , which leaves only infinitely many more numbers to check).
Writing proofs is a bit of an art. Like any art, to be truly great at it, you
need some sort of inspiration, as well as some foundational technique.
Just as musicians can learn proper fingering, and painters can learn the
proper way to hold a brush, we can look at the proper way to construct
arguments. A good place to start might be to study a classic.
Theorem 3.2.1 There are infinitely many primes.
Proof. Suppose this were not the case. That is, suppose there are only
finitely many primes. Then there must be a last, largest prime, call it p.
Consider the number
N p! + 1 (p · (p − 1) · · · · 3 · 2 · 1) + 1.
Now N is certainly larger than p. Also, N is not divisible by any
number less than or equal to p, since every number less than or equal to
p divides p!. Thus the prime factorization of N contains prime numbers
(possibly just N itself) all greater than p. So p is not the largest prime, a
contradiction. Therefore there are infinitely many primes.
qed
This proof is an example of a proof by contradiction, one of the standard
styles of mathematical proof. First and foremost, the proof is an argument. It contains sequence of statements, the last being the conclusion
which follows from the previous statements. The argument is valid so
the conclusion must be true if the premises are true. Let’s go through the
proof line by line.
1. Suppose there are only finitely many primes. [this is a premise. Note
the use of “suppose.”]
2. There must be a largest prime, call it p. [follows from line 1, by the
definition of “finitely many.”]
3. Let N p! + 1. [basically just notation, although this is the inspired part
of the proof; looking at p! + 1 is the key insight.]
4. N is larger than p. [by the definition of p!]
3.2. Proofs
215
5. N is not divisible by any number less than or equal to p. [by definition,
p! is divisible by each number less than or equal to p, so p! + 1 is not.]
6. The prime factorization of N contains prime numbers greater than
p. [since N is divisible by each prime number in the prime factorization of
N, and by line 5.]
7. Therefore p is not the largest prime. [by line 6, N is divisible by a prime
larger than p.]
8. This is a contradiction. [from line 2 and line 7: the largest prime is p
and there is a prime larger than p.]
9. Therefore there are infinitely many primes. [from line 1 and line 8:
our only premise lead to a contradiction, so the premise is false.]
We should say a bit more about the last line. Up through line 8, we have
a valid argument with the premise “there are only finitely many primes”
and the conclusion “there is a prime larger than the largest prime.” This
is a valid argument as each line follows from previous lines. So if the
premises are true, then the conclusion must be true. However, the conclusion is NOT true. The only way out: the premise must be false.
The sort of line-by-line analysis we did above is a great way to really
understand what is going on. Whenever you come across a proof in a
textbook, you really should make sure you understand what each line is
saying and why it is true. Additionally, it is equally important to understand the overall structure of the proof. This is where using tools from
logic is helpful. Luckily there are a relatively small number of standard
proof styles that keep showing up again and again. Being familiar with
these can help understand proof, as well as give ideas of how to write
your own.
Direct Proof
The simplest (from a logic perspective) style of proof is a direct proof.
Often all that is required to prove something is a systematic explanation
of what everything means. Direct proofs are especially useful when
proving implications. The general format to prove P → Q is this:
Assume P. Explain, explain, . . . , explain. Therefore Q.
Often we want to prove universal statements, perhaps of the form ∀x(P(x) →
Q(x)). Again, we will want to assume P(x) is true and deduce Q(x). But
what about the x? We want this to work for all x. We accomplish this by
fixing x to be an arbitrary element (of the sort we are interested in).
Here are a few examples. First, we will set up the proof structure for
a direct proof, then fill in the details.
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3. Symbolic Logic and Proofs
Example 3.2.2
Prove: For all integers n, if n is even, then n 2 is even.
Solution. The format of the proof with be this: Let n be an arbitrary
integer. Assume that n is even. Explain explain explain. Therefore
n 2 is even.
To fill in the details, we will basically just explain what it means
for n to be even, and then see what that means for n 2 . Here is a
complete proof.
Proof. Let n be an arbitrary integer. Suppose n is even. Then n 2k
for some integer k. Now n 2 (2k)2 4k 2 2(2k 2 ). Since 2k 2 is an
integer, n 2 is even.
qed
Example 3.2.3
Prove: For all integers a, b, and c, if a|b and b|c then a|c. Here x| y,
read “x divides y” means that y is a multiple of x (so x will divide
into y without remainder).
Solution. Even before we know what the divides symbol means,
we can set up a direct proof for this statement. It will go something
like this: Let a, b, and c be arbitrary integers. Assume that a|b and
b|c. Dot dot dot. Therefore a|c.
How do we connect the dots? We say what our hypothesis (a|b
and b|c) really means and why this gives us what the conclusion
(a|c) really means. Another way to say that a|b is to say that b ka
for some integer k (that is, that b is a multiple of a). What are we
going for? That c la, for some integer l (because we want c to be
a multiple of a). Here is the complete proof.
Proof. Let a, b, and c be integers. Assume that a|b and b|c. In other
words, b is a multiple of a and c is a multiple of b. So there are
integers k and j such that b ka and c jb. Combining these
(through substitution) we get that c jka. But jk is an integer, so
this says that c is a multiple of a. Therefore a|c.
qed
Proof by Contrapositive
Recall that an implication P → Q is logically equivalent to its contrapositive ¬Q → ¬P. There are plenty of examples of statements which
are hard to prove directly, but whose contrapositive can easily be proved
directly. This is all that proof by contrapositive does. It gives a direct
3.2. Proofs
217
proof of the contrapositive of the implication. This is enough because the
contrapositive is logically equivalent to the original implication.
The skeleton of the proof of P → Q by contrapositive will always look
roughly like this:
Assume ¬Q. Explain, explain, . . . explain. Therefore ¬P.
As before, if there are variables and quantifiers, we set them to be arbitrary
elements of our domain. Here are a couple examples:
Example 3.2.4
Is the statement “for all integers n, if n 2 is even, then n is even”
true?
Solution. This is the converse of the statement we proved above
using a direct proof. From trying a few examples, this statement
definitely appears this is true. So let’s prove it.
A direct proof of this statement would require fixing an arbitrary n and assuming that n 2 is even. But it is not at all clear
how this would allow us to conclude anything about n. Just because n 2 2k does not in itself suggest how we could write n as a
multiple of 2.
Try something else: write the contrapositive of the statement.
We get, for all integers n, if n is odd then n 2 is odd. This looks
much more promising. Our proof will look something like this:
Let n be an arbitrary integer. Suppose that n is not even. This
means that . . . . In other words . . . . But this is the same as saying
. . . . Therefore n 2 is not even.
Now we fill in the details:
Proof. We will prove the contrapositive. Let n be an arbitrary
integer. Suppose that n is not even, and thus odd. Then n 2k + 1
for some integer k. Now n 2 (2k +1)2 4k 2 +4k +1 2(2k 2 +2k)+1.
Since 2k 2 + 2k is an integer, we see that n 2 is odd and therefore not
even.
qed
Example 3.2.5
Prove: for all integers a and b, if a + b is odd, then a is odd or b is
odd.
Solution. The problem with trying a direct proof is that it will be
hard to separate a and b from knowing something about a + b. On
the other hand, if we know something about a and b separately,
then combining them might give us information about a + b. The
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3. Symbolic Logic and Proofs
contrapositive of the statement we are trying to prove is: for all
integers a and b, if a and b are even, then a + b is even. Thus our
proof will have the following format:
Let a and b be integers. Assume that a and b are both even. la
la la. Therefore a + b is even.
Here is a complete proof:
Proof. Let a and b be integers. Assume that a and b are even. Then
a 2k and b 2l for some integers k and l. Now a + b 2k + 2l 2(k +1). Since k + l is an integer, we see that a + b is even, completing
the proof.
qed
Note that our assumption that a and b are even is really the
negation of a or b is odd. We used De Morgan’s law here.
We have seen how to prove some statements in the form of implications: either directly or by contrapositive. Some statements are not
written as implications to begin with.
Example 3.2.6
Consider the statement, for every prime number p, either p 2 or
p is odd. We can rephrase this: for every prime number p, if p , 2,
then p is odd. Now try to prove it.
Solution.
Proof. Let p be an arbitrary prime number. Assume p is not odd.
So p is divisible by 2. Since p is prime, it must have exactly two
divisors, and it has 2 as a divisor, so p must be divisible by only 1 and
2. Therefore p 2. This completes the proof (by contrapositive).
qed
Proof by Contradiction
There might be statements
which really cannot be rephrased as implica√
tions. For example, “ 2 is irrational.” In this case, it is hard to know
where to start. What can we assume? Well, say we want to prove the
statement P. What if we could prove that ¬P → Q where Q was false? If
this implication is true, and Q is false, what can we say about ¬P? It must
be false as well, which makes P true!
This is why proof by contradiction works. If we can prove that ¬P
leads to a contradiction, then the only conclusion is that ¬P is false, so P
is true. That’s what we wanted to prove. In other words, if it is impossible
for P to be false, P must be true.
Here are a couple examples of proofs by contradiction:
3.2. Proofs
219
Example 3.2.7
√
Prove that 2 is irrational.
Solution.
√
Proof. Suppose not. Then 2 is equal to a fraction ba . Without loss
of generality, assume ba is in lowest terms (otherwise reduce the
fraction). So,
a2
2 2
b
2b 2 a 2
Thus a 2 is even, and as such a is even. So a 2k for some integer
k, and a 2 4k 2 . We then have,
2b 2 4k 2
b 2 2k 2
Thus b 2 is even, and as such b is even. Since a is also even,
√
we see that ba is not in lowest terms, a contradiction. Thus 2 is
irrational.
qed
Example 3.2.8
Prove: There are no integers x and y such that x 2 4y + 2.
Solution.
Proof. We proceed by contradiction. So suppose there are integers
x and y such that x 2 4y + 2 2(2y + 1). So x 2 is even. We have
seen that this implies that x is even. So x 2k for some integer
k. Then x 2 4k 2 . This in turn gives 2k 2 (2y + 1). But 2k 2 is
even, and 2y + 1 is odd, so these cannot be equal. Thus we have a
contradiction, so there must not be any integers x and y such that
x 2 4y + 2.
qed
Example 3.2.9
The Pigeonhole Principle: If more than n pigeons fly into n pigeon
holes, then at least one pigeon hole will contain at least two pigeons.
Prove this!
220
3. Symbolic Logic and Proofs
Solution.
Proof. Suppose, contrary to stipulation, that each of the pigeon
holes contain at most one pigeon. Then at most, there will be n
pigeons. But we assumed that there are more than n pigeons, so
this is impossible. Thus there must be a pigeonhole with more than
one pigeon.
qed
While we phrased this proof as a proof by contradiction, we
could have also used a proof by contrapositive since our contradiction was simply the negation of the hypothesis. Sometimes this
will happen, in which case you can use either style of proof. There
are examples however where the contradiction occurs “far away”
from the original statement.
Proof by (counter) Example
It is almost NEVER okay to prove a statement with just an example.
Certainly none of the statements proved above can be proved through an
example. This is because in each of those cases we are trying to prove that
something holds of all integers. We claim that n 2 being even implies that
n is even, no matter what integer n we pick. Showing that this works for
n 4 is not even close to enough.
This cannot be stressed enough. If you are trying to prove a statement
of the form ∀xP(x), you absolutely CANNOT prove this with an example.1
However, existential statements can be proven this way. If we want to
prove that there is an integer n such that n 2 − n + 41 is not prime, all we
need to do is find one. This might seem like a silly thing to want to prove
until you try a few values for n.
n2
n
− n + 41
1
41
2
43
3
47
4
53
5
61
6
71
7
83
So far we have gotten only primes. You might be tempted to conjecture,
“For all positive integers n, the number n 2 − n + 41 is prime.” If you
wanted to prove this, you would need to use a direct proof, a proof by
contrapositive, or another style of proof, but certainly it is not enough
to give even 7 examples. In fact, we can prove this conjecture is false by
proving its negation: “There is a positive integer n such that n 2 − n + 41 is
not prime.” Since this is an existential statement, it suffices to show that
there does indeed exist such a number.
In fact, we can quickly see that n 41 will give 412 which is certainly
not prime. You might say that this is a counterexample to the conjecture
1This is not to say that looking at examples is a waste of time. Doing so will often give
you an idea of how to write a proof. But the examples do not belong in the proof.
3.2. Proofs
221
that n 2 − n + 41 is always prime. Since so many statements in mathematics
are universal, making their negations existential, we can often prove that
a statement is false (if it is) by providing a counterexample.
Example 3.2.10
Above we proved, “for all integers a and b, if a + b is odd, then a is
odd or b is odd.” Is the converse true?
Solution. The converse is the statement, “for all integers a and b,
if a is odd or b is odd, then a + b is odd.” This is false! How do
we prove it is false? We need to prove the negation of the converse.
Let’s look at the symbols. The converse is
∀a∀b((O(a) ∨ O(b)) → O(a + b)).
We want to prove the negation:
¬∀a∀b((O(a) ∨ O(b)) → O(a + b)).
Simplify using the rules from the previous sections:
∃a∃b((O(a) ∨ O(b)) ∧ ¬O(a + b)).
As the negation passed by the quantifiers, they changed from ∀
to ∃. We then needed to take the negation of an implication, which
is equivalent to asserting the if part and not the then part.
Now we know what to do. To prove that the converse is false
we need to find two integers a and b so that a is odd or b is odd,
but a + b is not odd (so even). That’s easy: 1 and 3. (remember,
“or” means one or the other or both). Both of these are odd, but
1 + 3 4 is not odd.
Proof by Cases
We could go on and on and on about different proof styles (we haven’t
even mentioned induction or combinatorial proofs here), but instead we
will end with one final useful technique: proof by cases. The idea is
to prove that P is true by proving that Q → P and ¬Q → P for some
statement Q. So no matter what, whether or not Q is true, we know that
P is true. In fact, we could generalize this. Suppose we want to prove P.
We know that at least one of the statements Q1 , Q 2 , . . . , Q n are true. If
we can show that Q 1 → P and Q 2 → P and so on all the way to Q n → P,
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3. Symbolic Logic and Proofs
then we can conclude P. The key thing is that we want to be sure that one
of our cases (the Q i ’s) must be true no matter what.
If that last paragraph was confusing, perhaps an example will make
things better.
Example 3.2.11
Prove: For any integer n, the number (n 3 − n) is even.
Solution. It is hard to know where to start this, because we don’t
know much of anything about n. We might be able to prove that
n 3 −n is even if we knew that n was even. In fact, we could probably
prove that n 3 − n was even if n was odd. But since n must either
be even or odd, this will be enough. Here’s the proof.
Proof. We consider two cases: if n is even or if n is odd.
Case 1: n is even. Then n 2k for some integer k. This give
n 3 − n 8k 3 − 2k
2(4k 2 − k),
and since 4k 2 − k is an integer, this says that n 3 − n is even.
Case 2: n is odd. Then n 2k + 1 for some integer k. This gives
n 3 − n (2k + 1)3 − (2k + 1)
8k 3 + 6k 2 + 6k + 1 − 2k − 1
2(4k 3 + 3k 2 + 2k),
and since 4k 3 + 3k 2 + 2k is an integer, we see that n 3 − n is even
again.
Since n 3 − n is even in both exhaustive cases, we see that n 3 − n
is indeed always even.
qed